128 (2003)
MATHEMATICA BOHEMICA
No. 3, 237–252
CHARACTERIZATIONS OF THE 0-DISTRIBUTIVE SEMILATTICE
, Perundurai
(Received February 22, 2002)
Abstract. The 0-distributive semilattice is characterized in terms of semiideals, ideals
and filters. Some sufficient conditions and some necessary conditions for 0-distributivity
are obtained. Counterexamples are given to prove that certain conditions are not necessary
and certain conditions are not sufficient.
Keywords: semilattice, prime ideal, filter
MSC 2000 : 06A12, 06A99, 06B10, 06B99
1. Introduction and preliminaries
The 0-distributive lattice and the 0-distributive semilattice have been studied by
Varlet [7], [8], Pawar and Thakare [4], [5], Jayaram [3] and Balasubramani and
Venkatanarasimhan [1]. In this paper we obtain some characterizations of the 0distributive semilattice. For the lattice theoretic concepts which have now become
commonplace the reader is referred to Szasz [6] and Grätzer [2].
A semilattice is a partially ordered set in which any two elements have a greatest
lower bound. Let S be a semilattice. A semiideal of S is a nonempty subset A of S
such that a ∈ A, b 6 a (b ∈ S) ⇒ b ∈ A. An ideal of S is a semiideal A of S such
that the join of any finite number of elements of A, whenever it exists, belongs to A.
If a ∈ S, then {x ∈ S; x 6 a} is an ideal. It is called the principal ideal generated
by a and is denoted by (a]. A filter of S is a nonempty subset F of S such that (i)
a ∈ F , b > a (b ∈ S) ⇒ b ∈ F and (ii) a, b ∈ F ⇒ a ∧ b ∈ F . The dual of a principal
ideal is called a principal filter. The principal filter generated by a is denoted by [a).
A maximal ideal (filter) of S is a proper ideal (filter) which is not contained in any
other proper ideal (filter). A prime semiideal (ideal) is a proper semiideal (ideal)
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A such that a ∧ b ∈ A ⇒ a ∈ A or b ∈ A. A minimal prime semiideal (ideal) is
a prime semiideal (ideal) which does not contain any other prime semiideal (ideal).
Let F (S) denote the set of filters of S. A prime filter of S is a filter A such that B,
C ∈ F (S), B ∩ C ⊆ A, B ∩ C 6= ∅ ⇒ B ⊆ A or C ⊆ A. If A is a prime filter of S
and A1 , . . . , An ∈ F (S), A1 ∩ . . . ∩ An ⊆ A, A1 ∩ . . . ∩ An 6= ∅, then Ai ⊆ A for some
i ∈ {1, . . . , n}.
Let A be a nonempty subset of a semilattice S with 0, A∗ = {x ∈ S; a ∧ x = 0
for all a ∈ A} and A0 = {x ∈ S; a ∧ x = 0 for some a ∈ A}. Then A∗ is called the
annihilator of A and A0 is called the pseudoannihilator of A. If a ∈ S, we write (a)∗
for {a}∗ and (a)0 for {a}0 . We say that a is dense if (a)∗ = {0}. If sup(a)∗ ∈ (a)∗ , it
is called the pseudocomplement of a and is denoted by a∗ . A pseudocomplemented
semilattice is a semilattice with 0 in which every element has a pseudocomplement.
An ideal (semiideal) A of a semilattice S with 0 is said to be normal if A∗∗ = A.
The following five lemmas are contained in Venkatanarasimhan [9].
Lemma 1.1. The set I(S) of all ideals of a semilattice S forms a lattice under
set inclusion as the partial ordering relation. The meet in I(S) coincides with the
set intersection.
Lemma 1.2. Let S be a semilattice and {ai ; i ∈ I} any subset of S. Then
W T
T
ai
ai exists if and only if (ai ]
[ai ) is a principal ideal (principal filter).
W V
W T
V T
Whenever ai
ai exists then (ai ] =
ai
[ai ) =
ai .
V
Lemma 1.3. Let S be a semilattice. Then for a1 , . . . , an ∈ S, a1 ∨ . . . ∨ an exists
if and only if (a1 ] ∨ . . . ∨ (an ] is a principal ideal. Whenever a1 ∨ . . . ∨ an exists then
(a1 ] ∨ . . . ∨ (an ] = (a1 ∨ . . . ∨ an ].
Lemma 1.4. If {Ai ; i ∈ I} is a family of ideals of a semilattice, then
S x; (x] ⊆ (ai1 ] ∨ . . . ∨ (ain ]; ai1 , . . . , ain ∈ Ai .
W
Ai =
Lemma 1.5. Every proper filter of a semilattice with 0 is contained in a maximal
filter.
The following lemma is easily proved.
Lemma 1.6. Let A be a nonempty subset of a semilattice S with 0 and x ∈ S.
Then A∗ and A0 are semiideals of S and (x]∗ = [x)0 = (x)0 = (x)∗ .
The following four lemmas are contained in Venkatanarasimhan [10].
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Lemma 1.7. Let A be a nonempty proper subset of a semilattice S with 0. Then
A is a filter if and only if S − A is a prime semiideal.
Lemma 1.8. Let A be a nonempty subset of a semilattice S with 0. Then A is
a maximal filter if and only if S − A is a minimal prime semiideal.
Lemma 1.9. Any prime semiideal of a semilattice with 0 contains a minimal
prime semiideal.
Lemma 1.10. Let A be a nonempty subset of a semilattice with 0. Then A∗ is
the intersection of all minimal prime semiideals not containing A.
The following lemma is contained in Pawar and Thakare [4].
Lemma 1.11. Let A be a proper filter of a semilattice S with 0. Then A is
maximal if and only if for each x in S − A, there is some a in A such that a ∧ x = 0.
Lemma 1.12. Let A and B be filters of a semilattice S with 0 such that A and
B are disjoint. Then there is a minimal prime semiideal containing B 0 and disjoint
from A.
0
. It is easily seen that A ∨ B is a proper filter of S. Hence by Lemma 1.5,
A ∨ B ⊆ M for some maximal filter M . Now B ⊆ M and so M ∩ B 0 = ∅. By
Lemma 1.8, S − M is a minimal prime semiideal. Clearly B 0 ⊆ S − M and (S −
M ) ∩ A = ∅.
Lemma 1.13. Let A be a filter of a semilattice S with 0. Then A0 is the
intersection of all minimal prime semiideals disjoint from A.
. Let N be any minimal prime semiideal disjoint from A. If x ∈ A0 , then
x ∧ a = 0 for some a ∈ A and so x ∈ N .
Let y ∈ S − A0 . Then a ∧ y 6= 0 for all a ∈ A. Hence A ∨ [y) 6= S. By Lemma 1.5,
A ∨ [y) ⊆ M for some maximal filter M . By Lemma 1.8, S − M is a minimal prime
semiideal. Clearly (S − M ) ∩ A = ∅ and y ∈
/ S − M.
Lemma 1.14. Let S be a semilattice with 0. Then the set complement of a prime
filter is a prime ideal. If S is finite, then the set complement of a prime ideal is a
prime filter.
. Let A be a prime filter of S. By Lemma 1.7, S − A is a prime semiideal.
Let x1 , . . . , xn ∈ S − A and suppose x1 ∨ . . . ∨ xn exists. Since A is prime it follows
that x1 ∨ . . . ∨ xn ∈ S − A. Thus S − A is a prime ideal.
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Let S be finite and let A be any prime ideal of S. By Lemma 1.7, S − A is a filter.
Since S is finite, every filter of S is principal. Let a, b ∈ A be such that [a) ∩ [b) 6= ∅.
Let [a) ∩ [b) = {c1 , . . . , cn } and c = c1 ∧ . . . ∧ cn . Then c > a, b. If d > a, b then
d = cj for some j and so d > c. Thus c = a ∨ b ∈ A. Hence [a) ∩ [b) = [a ∨ b) S − A
proving S − A is prime.
2. Definition and characterizations
Definition 2.1. A 0-distributive lattice is a lattice with 0 in which a ∧ b = 0 =
a ∧ c implies a ∧ (b ∨ c) = 0.
Varlet [7], has proved that a lattice L bounded below is 0-distributive if and
only if the ideal lattice I(L) is pseudocomplemented. He also observed that for an
ideal lattice, the two notions of pseudocomplementedness and 0-distributivity are
equivalent. These results motivate the following definition.
Definition 2.2. A 0-distributive semilattice is a semilattice S with 0 such that
I(S), the lattice of ideals of S, is 0-distributive.
Theorem 2.3. Let S be a semilattice with 0. Then the following statements are
equivalent:
1. S is 0-distributive.
2. If A, A1 , . . . , An are ideals of S such that A ∩ A1 = . . . = A ∩ An = (0], then
A ∩ (A1 ∨ . . . ∨ An ) = (0].
3. If a, a1 , . . . , an are elements of S such that (a] ∩ (a1 ] = . . . = (a] ∩ (an ] = (0],
then (a] ∩ ((a1 ] ∨ . . . ∨ (an ]) = (0].
4. If M is a maximal filter of S, then S − M is a minimal prime ideal.
5. Every minimal prime semiideal of S is a minimal prime ideal.
6. Every prime semiideal of S contains a minimal prime ideal.
7. Every proper filter of S is disjoint from a minimal prime ideal.
8. For each nonzero element a of S, there is a minimal prime ideal not containing a.
9. For each nonzero element a of S, there is a prime ideal not containing a.
. 1 ⇒ 2: Suppose 1 holds and let A, A1 , . . . , An ∈ I(S) be such that
A ∩ A1 = . . . = A ∩ An = (0]. By 1, I(S) is 0-distributive. Hence A ∩ (A1 ∨ A2 ) = (0].
Assume A ∩ (A1 ∨ . . . ∨ Ak−1 ) = (0] for 2 < k 6 n. Then A ∩ (A1 ∨ . . . ∨ Ak−1 ∨ Ak ) =
A ∩ (B ∨ Ak ) where B = A1 ∨ . . . ∨ Ak−1 . By our induction hypothesis A ∩ B = (0].
Also A ∩ Ak = (0]. Consequently A ∩ (A1 ∨ . . . ∨ Ak ) = A ∩ (B ∨ Ak ) = (0]. Thus
the result follows by induction.
Obviously 2 ⇒ 3 and 8 ⇒ 9.
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3 ⇒ 1: Suppose 3 holds. Let A, B, C ∈ I(S) be such that A ∩ B = (0] = A ∩ C.
Then (a] ∩ (b] = (0] = (a] ∩ (c] for all a ∈ A, b ∈ B and c ∈ C. Let x ∈ A ∩ (B ∨ C).
Then x ∈ B ∨C. Hence (x] ⊆ (b1 ]∨. . .∨(bm ]∨(c1 ]∨. . .∨(cn ] for some b1 , . . . , bm ∈ B
and c1 , . . . , cn ∈ C. Also x ∈ A. Consequently (x] ∩ (bi ] = (0] for i = 1, . . . , m and
(x] ∩ (cj ] = (0] for j = 1, . . . , n. By 3, (x] ∩ ((b1 ] ∨ . . . ∨ (bm ] ∨ (c1 ] ∨ . . . ∨ (cn ]) = (0].
It follows that x = 0. Thus A ∩ (B ∨ C) = (0].
3 ⇒ 4: Suppose 3 holds. Let M be any maximal filter of S. By Lemma 1.8, S − M
is a minimal prime semiideal. Let x1 , . . . , xn ∈ S −M be such that x1 ∨. . .∨xn exists.
By Lemma 1.11, a1 ∧ x1 = . . . = an ∧ xn = 0 for some a1 , . . . , an ∈ M . Let a =
a1 ∧. . .∧an . Then a ∈ M and a∧xi = 0 for i = 1, . . . , n. By Lemma 1.2, (a]∩(xi ] = (0]
for i = 1, . . . , n. By Lemma 1.3, (a] ∩ (x1 ∨ . . . ∨ xn ] = (a] ∩ ((x1 ] ∨ . . . ∨ (xn ]) = (0]
by 3. It follows that a ∧ (x1 ∨ . . . ∨ xn ) = 0. Hence x1 ∨ . . . ∨ xn ∈ S − M . Thus
S − M is an ideal.
4 ⇒ 5: Suppose 4 holds. Let N be any minimal prime semiideal of S. By
Lemma 1.8, S − N is a maximal filter. By 4, N = S − (S − N ) is a minimal prime
ideal.
5 ⇒ 6: Suppose 5 holds and let A be any prime semiideal of S. By Lemma 1.9,
A ⊇ N for some minimal prime semiideal N . By 5, N is a minimal prime ideal.
6 ⇒ 7: Suppose 6 holds and let A be any proper filter of S. By Lemma 1.7,
S − A is a prime semiideal. By 6, S − A contains a minimal prime ideal N . Clearly
A ∩ N = ∅.
7 ⇒ 8: Suppose 7 holds and let a be any nonzero element of S. By 7, [a) is disjoint
from a minimal prime ideal N . Clearly a ∈
/ N.
9 ⇒ 3: Suppose 9 holds. Let a, a1 , . . . , an ∈ S such that (a] ∩ (a1 ] = . . . = (a] ∩
(an ] = (0] and (a]∩((a1 ]∨. . .∨(an ]) 6= (0]. Then there exists x ∈ (a]∩((a1 ]∨. . .∨(an ])
such that x 6= 0. By 9 there is a prime ideal A such that x ∈
/ A. By Lemma 1.7,
S − A is a proper filter and clearly a ∈ S − A. Consequently a1 , . . . , an ∈ A. It
follows that (a1 ] ∨ . . . ∨ (an ] ⊆ A and so x ∈ A. Thus we get a contradiction. Hence
(a] ∩ (a1 ] = . . . = (a] ∩ (an ] = (0] ⇒ (a] ∩ ((a1 ] ∨ . . . ∨ (an ]) = (0].
Theorem 2.4. Let S be a semilattice with 0. Then the following statements are
equivalent:
1. S is 0-distributive.
2. If A is a nonempty subset of S and B is a proper filter intersecting A, there is
a minimal prime ideal containing A∗ and disjoint from B.
3. If A is a nonempty subset of S and B is a proper filter intersecting A, there is
a prime ideal containing A∗ and disjoint from B.
4. If A is a nonempty subset of S and B is a prime semiideal not containing A,
there is a minimal prime ideal containing A∗ and contained in B.
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5. If A is a nonempty subset of S and B is a prime semiideal not containing A,
there is a prime ideal containing A∗ and contained in B.
6. For each nonzero element a of S and each proper filter B containing a, there is
a prime ideal containing (a)∗ and disjoint from B.
7. For each nonzero element a of S and each prime semiideal B not containing a,
there is a prime ideal containing (a)∗ and contained in B.
8. If A and B are filters of S such that A and B 0 are disjoint, there is a minimal
prime ideal containing B 0 and disjoint from A.
9. If A and B are filters of S such that A and B 0 are disjoint, there is a prime
ideal containing B 0 and disjoint from A.
10. If A is a filter of S and B is a prime semiideal containing A0 , there is a minimal
prime ideal containing A0 and contained in B.
11. If A is a filter of S and B is a prime semiideal containing A0 , there is a prime
ideal containing A0 and contained in B.
12. For each nonzero element a in S and each filter A disjoint from (a)∗ , there is a
prime ideal containing (a)∗ and disjoint from A.
13. For each nonzero element a in S and each prime semiideal B containing (a)∗ ,
there is a prime ideal containing (a)∗ and contained in B.
. 1 ⇒ 2: Suppose 1 holds. Let A be a nonempty subset of S and B any
proper filter such that B ∩ A 6= ∅. By Lemma 1.7, S − B is a prime semiideal and by
Lemma 1.9, S − B ⊇ N for some minimal prime semiideal N . Clearly N ∩ B = ∅.
Also S − B 6⊇ A and so N 6⊇ A. By Lemma 1.10, N ⊇ A∗ . Since S is 0-distributive,
N is a minimal prime ideal [see Theorem 2.3, 5].
By Lemma 1.7, it follows that 2 ⇒ 4, 3 ⇒ 5, 8 ⇒ 10, 9 ⇒ 11 and 12 ⇒ 13.
Obviously 2 ⇒ 3, 2 ⇒ 6, 4 ⇒ 5, 4 ⇒ 7, 8 ⇒ 9, 10 ⇒ 11 ⇒ 13 and 5 ⇒ 7.
1 ⇒ 8: Suppose 1 holds. Let A and B be filters of S such that A ∩ B 0 6= ∅. By
Lemma 1.12, there is a minimal prime semiideal N such that N ⊇ B 0 and N ∩A = ∅.
Since S is 0-distributive it follows that N is a minimal prime ideal [see Theorem 2.3,
5].
8 ⇒ 12: By Lemma 1.6, (x)∗ = [x)0 for all x ∈ S. Hence the result.
6 ⇒ 1: Suppose 6 holds. Let a be any nonzero element of S. Now [a) is a proper
filter containing a. By 6, there is a prime ideal N containing (a)∗ and disjoint from
[a). Clearly a ∈
/ N . Thus S is 0-distributive [see Theorem 2.3, 9].
7 ⇒ 1: Suppose 7 holds. Let a be any nonzero element of S. Now S − [a) is a
prime semiideal not containing a. By 7 there is a prime ideal N containing (a)∗ and
contained in S − [a). Clearly a ∈
/ N . Thus S is 0-distributive [See Theorem 2.3, 9].
13 ⇒ 1: Suppose 13 holds and let a be any nonzero element of S. By Lemma 1.7,
S − [a) is a prime semiideal not containing a. Since (a) ∩ (a)∗ = (0] ⊆ S − [a) it
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follows that S − [a) contains (a)∗ . By 13, there is a prime ideal N containing (a)∗
and contained in S − [a). Clearly a ∈ N . Thus S is 0-distributive [see Theorem 2.3,
9].
Theorem 2.5. Let S be a semilattice with 0. Then the following statements are
equivalent:
1. S is 0-distributive.
2. For any nonempty subset A of S, A∗ is the intersection of all minimal prime
ideals not containing A.
3. For any filter A of S, A0 is the intersection of all minimal prime ideals disjoint
from A.
4. For each a in S, (a)∗ is an ideal.
5. Every normal semiideal of S is an intersection of minimal prime ideals.
6. For any finite number of ideals A, A1 , . . . , An of S,
(A ∩ (A1 ∨ . . . ∨ An ))∗ = (A ∩ A1 )∗ ∩ . . . ∩ (A ∩ An )∗ .
7. For any three ideals A, B, C of S,
(A ∩ (B ∨ C))∗ = (A ∩ B)∗ ∩ (A ∩ C)∗ .
8. For any finite number of ideals A, A1 , . . . , An of S,
((A ∨ A1 ) ∩ . . . ∩ (A ∨ An ))∗ = A∗ ∩ (A1 ∩ . . . An )∗ .
9. For any three ideals A, B, C of S,
((A ∨ B) ∩ (A ∨ C))∗ = A∗ ∩ (B ∩ C)∗ .
10. For any finite number of elements a, a1 , . . . , an of S,
((a] ∩ ((a1 ] ∨ . . . ∨ (an ]))∗ = ((a] ∩ (a1 ])∗ ∩ . . . ∩ ((a] ∩ (an ])∗ .
11. For any finite number of elements a1 , . . . , an of S,
((a1 ] ∨ . . . ∨ (an ])∗ = (a1 ]∗ ∩ . . . ∩ (an ]∗ .
12. I(S) is pseudocomplemented.
. 1 ⇒ 2: Follows by Lemma 1.10 and Theorem 2.3, 5.
1 ⇒ 3: Follows by Lemma 1.13 and Theorem 2.3, 5.
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3 ⇒ 4: By Lemma 1.6, (a)∗ = [a)0 . Hence the result.
4 ⇒ 1: Suppose 4 holds. Let a, a1 , . . . , an ∈ S be such that (a] ∩ (a1 ] = . . . =
(a] ∩ (an ] = (0]. Then a1 , . . . , an ∈ (a)∗ . By 4 it follows that (a1 ] ∨ . . . ∨ (an ] ⊆ (a)∗ .
Hence (a] ∩ ((a1 ] ∨ . . . ∨ (an ]) = (0]. Thus S is 0-distributive [see Theorem 2.3, 3].
Obviously 6 ⇒ 7, 8 ⇒ 9 and 6 ⇒ 10.
2 ⇒ 5: Suppose 2 holds. Let A be any normal semiideal of S. Then A = B ∗
for some semiideal B. By 2, B ∗ is the intersection of all minimal prime ideals not
containing B. Hence the result.
5 ⇒ 4: By Lemma 1.6, (a)∗ = (a]∗ for all a ∈ S. Hence the result.
2 ⇒ 6: Suppose 2 holds. Let A, A1 , . . . , An ∈ I(S). If Q is any minimal prime
ideal of S such that Q A∩(A1 ∨. . .∨An ), then Q A∩Aj for some j ∈ {1, . . . , n}.
By 2 it follows that (A ∩ (A1 ∨ . . . ∨ An ))∗ ⊇ (A ∩ A1 )∗ ∩ . . . ∩ (A ∩ An )∗ . The reverse
inclusion is obvious.
7 ⇒ 1: Suppose 7 holds. Then for A, B, C ∈ I(S) we have (A ∩ (B ∨ C))∗ =
(A ∩ B)∗ ∩ (A ∩ C)∗ . By replacing A by B ∨ C it follows that (B ∨ C)∗ = B ∗ ∩ C ∗ .
Suppose A ∩ B = (0] = A ∩ C. Then (a] ∩ (b] = (0] = (a] ∩ (c] for all a ∈ A, b ∈ B
and c ∈ C. Hence a ∈ B ∗ ∩ C ∗ for all a ∈ A. Hence a ∈ (B ∨ C)∗ . Consequently
A ⊆ (B ∨ C)∗ . It follows that A ∩ (B ∨ C) = (0].
2 ⇒ 8: Suppose 2 holds, let A, A1 , . . . , An be ideals of S and let Q be any minimal
prime ideal such that Q (A ∨ A1 ) ∩ . . . ∩ (A ∨ An ). Then Q A ∨ A1 , . . . , A ∨ An
and so Q A or Q A1 ∩. . . ∩An . By 2 it follows that ((A∨A1 )∩. . . ∩(A∨An ))∗ ⊇
A∗ ∩ (A1 ∩ . . . ∩ An )∗ . The reverse inclusion is obvious.
9 ⇒ 1: Suppose 9 holds. Then for any three ideals A, B, C of S, ((A ∨ B) ∩
(A ∨ C))∗ = A∗ ∩ (B ∩ C)∗ . By replacing C by B and A by C it follows that
(B ∨ C)∗ = B ∗ ∩ C ∗ . Suppose A ∩ B = (0] = A ∩ C. Then (a] ∩ (b] = (0] = (a] ∩ (c]
for all a ∈ A, b ∈ B and c ∈ C. Hence a ∈ B ∗ ∩ C ∗ for all a ∈ A. Hence a ∈ (B ∨ C)∗
for all a ∈ A. Consequently A ⊆ (B ∨ C)∗ . It follows that A ∩ (B ∨ C) = (0]. Thus
S is 0-distributive.
10 ⇒ 1: Suppose 10 holds. Let a, a1 , . . . , an ∈ S such that (a] ∩ (a1 ] = . . . =
(a] ∩ (an ] = (0]. Then ((a] ∩ (a1 ])∗ = . . . = ((a] ∩ (an ])∗ = S. Hence ((a] ∩ (a1 ])∗ ∩
. . . ∩ ((a] ∩ (an ])∗ = S. By 10, ((a] ∩ ((a1 ] ∨ . . . ∨ (an ]))∗ = S. Consequently
(a] ∩ ((a1 ] ∨ . . . ∨ (an ]) = (0]. It follows that S is 0-distributive [see Theorem 2.3, 3].
6 ⇒ 11: Suppose 6 holds. Then for any finite number of ideals A, A1 , . . . , An of
S, (A ∩ (A1 ∨ . . . ∨ An ))∗ = (A ∩ A1 )∗ ∩ . . . ∩ (A ∩ An )∗ . By taking A = A1 ∨ . . . ∨ An
it follows that (A1 ∨ . . . ∨ An )∗ = A∗1 ∩ . . . ∩ A∗n . Hence the result.
11 ⇒ 1: Suppose 11 holds. Let a, a1 , . . . , an ∈ S be such that (a] ∩ (a1 ] = . . . =
(a]∩(an ] = (0]. Then a ∈ (a1 ]∗ ∩. . .∩(an ]∗ . By 11 it follows that a ∈ ((a1 ]∨. . .∨(an ])∗ .
Hence (a] ∩ ((a1 ] ∨ . . . ∨ (an ]) = (0]. Thus S is 0-distributive [see Theorem 2.3, 3].
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2 ⇒ 12: Suppose 2 holds. Let A ∈ I(S). Then by 2 it follows that A∗ is an ideal.
If B ∈ I(S) is such that A ∩ B = (0] and x ∈ B, then a ∧ x = 0 for all a ∈ A and so
x ∈ A∗ . Thus B ⊆ A∗ . It follows that A∗ is the pseudocomplement of A.
12 ⇒ 1: Suppose 12 holds. Then every principal ideal of S has a pseudocomplement in I(S). Let a, a1 , . . . , an ∈ S be such that (a] ∩ (a1 ] = . . . = (a] ∩ (an ] = (0].
Then (ai ] ⊆ (a]∗ for i = 1, . . . , n and so ((a1 ] ∨ . . . ∨ (an ]) ⊆ (a]∗ . Consequently
(a] ∩ ((a1 ] ∨ . . . ∨ (an ]) = (0]. Thus S is 0-distributive [see Theorem 2.3, 3].
!"
2.6. According to Varlet [8], an ideal of a semilattice S is a nonempty
subset I of S such that (i) y 6 x and x ∈ I imply y ∈ I; (ii) for any x, y ∈ I there
exists a z ∈ I such that z > x and z > y. According to him a semilattice S with 0
is said to be 0-distributive if for any a ∈ S, the subset (a)∗ = {x ∈ S; x ∧ a = 0} is
an ideal.
Let S be a 0-distributive semilattice in Varlet’s sense. Then for each a ∈ S, (a)∗
is a Varlet ideal and therefore an ideal in our sense. Thus S is 0-distributive in our
sense. The converse is not true. Consider the semilattice S = {0, a, b, c} in which
the ordering is defined by 0 < a, b, c; akb; akc; and bkc. Clearly S is 0-distributive in
our sense but not in Varlet’s sense.
We give below some additional characterizations when the semilattice is finite.
Theorem 2.7. Let S be a finite semilattice. Then the following statements are
equivalent:
1. S is 0-distributive.
2. If a, b, c are elements of S such that (a]∩(b] = (0] = (a]∩(c] then (a]∩((b]∨(c]) =
(0].
3. Every maximal filter of S is prime.
4. Each nonzero element of S is contained in a prime filter.
5. If A is a nonempty subset of S and B is a proper filter intersecting A, there is
a prime filter containing B and disjoint from A∗ .
6. If A is a nonempty subset of S and B is a prime semiideal not containing A,
there is a prime filter containing S − B and disjoint from A∗ .
7. For each nonzero element a of S and each proper filter B containing a, there is
a prime filter containing B and disjoint from (a)∗ .
8. For each nonzero element a of S and each prime semiideal B not containing a,
there is a prime filter containing S − B and disjoint from (a)∗ .
9. If A and B are filters of S such that A and B 0 are disjoint, there is a prime
filter containing A and disjoint from B 0 .
10. If A is a filter of S and B is a prime semiideal containing A0 , there is a prime
filter containing S − B and disjoint from A0 .
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11. For each nonzero element a in S and each filter A disjoint from (a)∗ , there is a
prime filter containing A and disjoint from (a)∗ .
12. For each nonzero element a in S and each prime semiideal B containing (a)∗ ,
there is a prime filter containing S − B and disjoint from (a)∗ .
. Obviously 1 ⇒ 2, 6 ⇒ 8, 10 ⇒ 12, 5 ⇒ 7 and 9 ⇒ 11.
2 ⇒ 1: Suppose 2 holds and let a, a1 , . . . , an ∈ S be such that (a] ∩ (a1 ] = . . . =
(a] ∩ (an ] = (0]. Let A = (a1 ] ∪ . . . ∪ (an ], let B = {b1 , . . . , bm } be the set of existing
suprema of nonempty subsets of A and b ∈ B. Then (a1 ]∨. . .∨(an ] = (b1 ]∪. . .∪(bm ]
and b = c1 ∨ . . . ∨ ck for some c1 , . . . , ck ∈ A. If p, q ∈ {1, . . . , k}, clearly b is an
upperbound of {cp , cq }. Thus the set C of upperbounds of {cp , cq } is nonempty and
inf C = cp ∨ cq . Also (a] ∩ (cp ] = (0] = (a] ∩ (cq ], so that (a] ∩ ((cp ] ∨ (cq ]) = (0]
by 2. It is easily seen that every nonempty subset of {c1 , . . . , ck } has a supremum
and by induction it follows that (a] ∩ (b] = (a] ∩ ((c1 ] ∨ . . . ∨ (ck ]) = (0]. Hence
(a] ∩ ((a1 ] ∨ . . . ∨ (an ]) = (a] ∩ ((b1 ] ∪ . . . ∪ (bm ]) = ((a] ∩ (b1 ]) ∪ . . . ∪ ((a] ∩ (bm ]) = (0].
Consequently S is 0-distributive [see Theorem 2.3, 3].
1 ⇒ 3: Suppose 1 holds. Let M be any maximal filter of S. Since S is finite,
every filter of S is principal. Let a, b ∈ S − M be such that [a) ∩ [b) 6= ∅. Let
[a) ∩ [b) = {c1 , . . . , cn } and c = c1 ∧ . . . ∧ cn . Then c > a, b as ci > a, b for all
i. If d ∈ S and d > a, b, then d = cj for some j, so that d > c. Thus c = a ∨ b.
Also S − M is an ideal [see Theorem 2.3, 4]. Hence a ∨ b ∈ S − M . It follows that
[a) ∩ [b) = [a ∨ b) M , proving M is prime.
3 ⇒ 4: Suppose 3 holds. Let a be any nonzero element of S. By Lemma 1.5, [a)
is contained in a maximal filter M . By 3, M is prime. Clearly a ∈ M .
4 ⇒ 1: Suppose 4 holds. Let a be any nonzero element of S. By 4, a ∈ B for
some prime filter B. By Lemma 1.14, S − B is a prime ideal and clearly a ∈
/ S − B.
It follows that S is 0-distributive [see Theorem 2.3, 9].
3 ⇒ 5: Suppose 3 holds. Let A be a nonempty subset of S and B a proper filter
such that B ∩ A 6= ∅. By Lemma 1.5, B ⊆ M for some maximal filter M . By 3, M is
prime. By Lemma 1.8, S − M is a minimal prime semiideal and clearly S − M A.
Hence S − M ⊇ A∗ and so M ∩ A∗ = ∅.
5 ⇒ 6: Suppose 5 holds. Let A be a nonempty subset of S and B a prime semiideal
such that B A. By Lemma 7, S − B is a proper filter and clearly (S − B) ∩ A 6= ∅.
By 5 there is a prime filter containing S − B and disjoint from A∗ .
7 ⇒ 8: Similar to 5 ⇒ 6.
8 ⇒ 1: Suppose 8 holds and let a be any nonzero element of S. Now S − [a)
is a prime semiideal not containing a. By 8 there is a prime filter N containing
S − (S − [a)) = [a) and disjoint from (a)∗ . By Lemma 1.14, S − N is a prime ideal
and clearly a ∈
/ S − N . Thus S is 0-distributive [see Theorem 2.3, 9].
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3 ⇒ 9: Suppose 3 holds. Let A and B be filters of S such that A and B 0 are
disjoint. By Lemma 1.12, there is a minimal prime semiideal N such that N ⊇ B 0
and N ∩ A = ∅. By Lemma 1.8, S − N is a maximal filter. Clearly S − N ⊇ A and
(S − N ) ∩ B 0 = ∅. By 3, S − N is prime.
9 ⇒ 10: Suppose 9 holds. Let A be a filter of S and B a prime semiideal such
that B ⊇ A0 . By Lemma 1.7, S − B is a proper filter and clearly (S − B) ∩ A0 = ∅.
By 9, there is a prime filter containing S − B and disjoint from A0 .
11 ⇒ 12: Similar to 5 ⇒ 6.
12 ⇒ 4: Suppose 12 holds. Let a be any nonzero element of S. Now S − [a) is a
prime semiideal not containing (a]. Since (a] ∩ (a]∗ = (0] ⊆ S − [a) it follows that
(a)∗ ⊆ S − [a). By 12 there is a prime filter N containing S − (S − [a)) = [a) and
disjoint from (a)∗ . Clearly a ∈ N .
Theorem 2.8. Let S be a finite semilattice. Then the following statements are
equivalent:
1. S is 0-distributive.
2. For any finite number of filters A, A1 , . . . , An of S such that A ∩ Ai 6= ∅ for all
i ∈ {1, . . . , n},
((A ∩ A1 ) ∨ . . . ∨ (A ∩ An ))0 = A0 ∩ (A1 ∨ . . . ∨ An )0 .
3. For any three filters A, B, C of S such that A ∩ B 6= ∅ and A ∩ C 6= ∅,
((A ∩ B) ∨ (A ∩ C))0 = A0 ∩ (B ∨ C)0 .
4. For all a, b, c in S such that [a) ∩ [b) 6= ∅ and [a) ∩ [c) 6= ∅,
(([a) ∩ [b)) ∨ ([a) ∩ [c)))0 = [a)0 ∩ ([b) ∨ [c))0 .
5. For any finite number of filters A, A1 , . . . , An of S such that A1 ∩ . . . ∩ An 6= ∅,
(A ∨ (A1 ∩ . . . ∩ An ))0 = (A ∨ A1 )0 ∩ . . . ∩ (A ∨ An )0 .
6. For any three filters A, B, C of S such that B ∩ C 6= ∅,
(A ∨ (B ∩ C))0 = (A ∨ B)0 ∩ (A ∨ C)0 .
7. For any finite number of elements a, a1 , . . . , an of S such that [a1 )∩. . .∩[an ) 6= ∅,
([a) ∨ ([a1 ) ∩ . . . ∩ [an )))0 = ([a) ∨ [a1 ))0 ∩ . . . ∩ ([a) ∨ [an ))0 .
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8. For all a, b, c in S, with [b) ∩ [c) 6= ∅,
([a) ∨ ([b) ∩ [c)))0 = ([a) ∨ [b))0 ∩ ([a) ∨ [c))0 .
9. For any finite number of elements a1 , . . . , an of S such that [a1 ) ∩ . . . ∩ [an ) 6= ∅,
([a1 ) ∩ . . . ∩ [an ))0 = [a1 )0 ∩ . . . ∩ [an )0 .
10. For all a, b in S with [a) ∩ [b) 6= ∅, ([a) ∩ [b))0 = [a)0 ∩ [b)0 .
11. For all a, b, c in S, ((a] ∩ ((b] ∨ (c]))∗ = ((a] ∩ (b])∗ ∩ ((a] ∩ (c])∗ .
12. For all a, b, c in S,
(((a] ∨ (b]) ∩ ((a] ∨ (c]))∗ = (a]∗ ∩ ((b] ∩ (c])∗ .
13. For all a, b in S, ((a] ∨ (b])∗ = (a]∗ ∩ (b]∗ .
. 1 ⇒ 2: Suppose 1 holds and let A, A1 , . . . , An be filters of S such that
A ∩ Ai 6= ∅ for all i ∈ {1, . . . , n}. If Q is any minimal prime ideal of S such that
Q ∩ ((A ∩ A1 ) ∨ . . . ∨ (A ∩ An )) = ∅, then Q ∩ (A ∩ A1 ) = . . . = Q ∩ (A ∩ An ) = ∅.
By Lemma 1.14, S − Q is a prime filter and S − Q ⊇ (A ∩ A1 ), . . . , (A ∩ An ). Hence
S − Q ⊇ A or S − Q ⊇ A1 ∨ . . . ∨ An and so Q ∩ A = ∅ or Q ∩ (A1 ∨ . . . ∨ An ) = ∅. It
follows that ((A ∩ A1 ) ∨ . . . ∨ (A ∩ An ))0 ⊇ A0 ∩ (A1 ∨ . . . ∨ An )0 [see Theorem 2.5,
3]. The reverse inclusion is obvious.
Obviously 2 ⇒ 3 ⇒ 4, 5 ⇒ 6 ⇒ 8 and 5 ⇒ 7 ⇒ 8.
4 ⇒ 10: Follows by taking c = b in 4.
1 ⇒ 5: Suppose 1 holds. Let A, A1 , . . . , An be filters of S such that A1 ∩. . .∩An 6=
∅. If Q is any minimal prime ideal of S such that Q ∩ (A ∨ (A1 ∩ . . . ∩ An )) = ∅, then
Q ∩ A = ∅ = Q ∩ (A1 ∩ . . . ∩ An ). By Lemma 1.14, S − Q is a prime filter and clearly
S − Q ⊇ A, A1 ∩ . . . ∩ An . Hence S − Q ⊇ A ∨ Aj and so Q ∩ (A ∨ Aj ) = ∅ for some
j ∈ {1, . . . , n}. It follows that (A ∨ (A1 ∩ . . . ∩ An ))0 ⊇ (A ∨ A1 )0 ∩ . . . ∩ (A ∨ An )0
[see Theorem 2.5, 3]. The reverse inclusion is obvious.
10 ⇒ 9: Suppose 10 holds and let a1 , . . . , an ∈ S be such that [a1 ) ∩ . . . ∩ [an ) 6= ∅.
Then ([a1 )∩[a2 ))0 = [a1 )0 ∩[a2 )0 . Assume ([a1 )∩. . .∩[ak−1 ))0 = [a1 )0 ∩. . .∩[ak−1 )0
for 2 < k 6 n. Let x ∈ [a1 )0 ∩ . . . ∩ [ak )0 . Then x ∈ [a1 )0 ∩ . . . ∩ [ak−1 )0 =
([a1 ) ∩ . . . ∩ [ak−1 ))0 by our induction hypothesis. Hence x ∧ y = 0 for some y ∈
([a1 ) ∩ . . . ∩ [ak−1 )). Thus x ∈ [y)0 ∩ [ak )0 = ([y) ∩ [ak ))0 ⊆ ([a1 ) ∩ . . . ∩ [ak ))0 so
that ([a1 )0 ∩ . . . ∩ [ak )0 ) ⊆ ([a1 ) ∩ . . . ∩ [ak ))0 . The reverse inclusion is obvious. By
induction it follows that ([a1 ) ∩ . . . ∩ [an ))0 = [a1 )0 ∩ . . . ∩ [an )0 .
9 ⇒ 1: Suppose 9 holds. Let a ∈ S and let a1 , . . . , an ∈ (a)∗ be such that
a1 ∨ . . . ∨ an exists. Then a ∧ a1 = . . . = a ∧ an = 0 and so a ∈ [a1 )0 ∩ . . . ∩ [an )0 =
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([a1 ) ∩ . . . ∩ [an ))0 by 9. That is a ∈ [a1 ∨ . . . ∨ an )0 . Hence a ∧ (a1 ∨ . . . ∨ an ) = 0,
so that a1 ∨ . . . ∨ an ∈ (a)∗ . Thus (a)∗ is an ideal. It follows that S is 0-distributive
[see Theorem 2.5, 4].
8 ⇒ 1: Suppose 8 holds and let a, b, c ∈ S such that (a] ∩ (b] = (0] = (a] ∩ (c].
Let X = {x1 , . . . , xn } be the set of existing suprema of nonempty subsets of (b] ∪ (c]
and x ∈ X. Then (b] ∨ (c] = (x1 ] ∪ . . . ∪ (xn ] and x = y1 ∨ . . . ∨ ym for some
y1 , . . . , ym ∈ (b] ∪ (c]. If p, q ∈ {1, . . . , m}, clearly x is an upperbound of {yp , yq }.
Thus the set Y of upperbounds of {yp , yq } is nonempty and inf Y = yp ∨ yq . Also
a ∧ yp = 0 = a ∧ yq . Hence ([a) ∨ [yp ))0 = S = ([a) ∨ [yq ))0 . Let z ∈ (a] ∩ ((yp ] ∨ (yq ]).
Then z 6 a and z 6 yp ∨ yq . Now z ∈ S = ([a) ∨ [yp ))0 ∩ ([a) ∨ [yq ))0 = ([a) ∨ ([yp ) ∩
[yq ))0 = ([a) ∨ [yp ∨ yq ))0 by 8, so that z ∧ t = 0 for some t ∈ [a) ∨ [yp ∨ yq ). Thus
z = z ∧ a ∧ (yp ∨ yq ) 6 z ∧ t = 0 and consequently (a] ∩ ((yp ] ∨ (yq ]) = (0]. It is easily
seen that every nonempty subset of {y1 , . . . , ym } has a supremum and by induction
it follows that (a] ∩ (x] = (a] ∩ ((y1 ] ∨ . . . ∨ (ym ]) = (0]. Hence (a] ∩ ((b] ∨ (c]) =
(a] ∩ ((x1 ] ∪ . . . ∪ (xn ]) = ((a] ∩ (x1 ]) ∪ . . . ∪ ((a] ∩ (xn ]) = (0]. Thus S is 0-distributive
[see Theorem 2.7, 2].
1 ⇒ 11: Suppose 1 holds. Then for all A, B, C ∈ I(S) we have (A ∩ (B ∨ C))∗ =
(A ∩ B)∗ ∩ (A ∩ C)∗ [see Theorem 2.5, 7]. Hence 11 follows.
1 ⇒ 12: Suppose 1 holds. Then for all A, B, C ∈ I(S) we have ((A ∨ B) ∩ (A ∨
C))∗ = A∗ ∩ (B ∩ C)∗ [see Theorem 2.5,9]. Hence 12 follows.
12 ⇒ 13: Follows by taking c = b in 12.
13 ⇒ 1: Suppose 13 holds. Let a, b, c ∈ S be such that (a] ∩ (b] = (0] = (a] ∩ (c].
Then a ∈ (b]∗ ∩ (c]∗ = ((b] ∨ (c])∗ by 13. Hence (a] ∩ ((b] ∨ (c]) = (0]. Thus S is
0-distributive [see Theorem 2.7, 2].
11 ⇒ 1: Suppose 11 holds. Let a, b, c ∈ S be such that (a] ∩ (b] = (0] = (a] ∩ (c].
Then ((a] ∩ (b])∗ ∩ ((a] ∩ (c])∗ = S. Hence By 11, ((a] ∩ ((b] ∨ (c]))∗ = S. It follows
that (a] ∩ ((b] ∨ (c]) = (0]. Thus S is 0-distributive [see Theorem 2.7, 2].
Theorem 2.9. Any one of the conditions 3 to 12 of Theorem 2.7 is sufficient for
a semilattice S with 0 (not necessarily finite) to be 0-distributive. These conditions
are also necessary in the case of a lattice.
. Suppose 3 of Theorem 2.7 holds and let M be any maximal filter of
S. By Lemma 1.8, S − M is a minimal prime semiideal. Let x1 , . . . , xn ∈ S − M
and suppose x1 ∨ . . . ∨ xn exists. By 3, M is prime and clearly [xi ) M for i =
1, . . . , n. Hence by Lemma 1.2, [x1 ∨ . . . ∨ xn ) = [x1 ) ∩ . . . ∩ [xn ) M . Consequently
x1 ∨ . . . ∨ xn ∈ S − M and so S − M is an ideal. It follows that S is 0-distributive
[see Theorem 2.3, 4].
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The sufficiency of the condition 4 of Theorem 2.7 follows by Lemma 1.14 and
Theorem 2.3 [see Theorem 2.3, 9]. The sufficiency of the conditions 5 to 12 of
Theorem 2.7 follows by Lemma 1.14 and Theorem 2.4 [see Theorem 2.4, 3, 5, 6, 7,
9, 11, 12, 13].
Theorem 2.10. Any one of the conditions 2 to 10 of Theorem 2.8 is sufficient for
a semilattice S with 0 (not necessarily finite) to be 0-distributive. These conditions
are also necessary in the case of a lattice.
. Obviously 2 ⇒ 3 ⇒ 4 and 5 ⇒ 6 ⇒ 8.
4 ⇒ 10: Follows by taking c = b in 4.
10 ⇒ 9: Same proof as in Theorem 2.8.
Suppose 9 holds. Let a ∈ S and let a1 , . . . , an ∈ (a)∗ be such that a1 ∨. . .∨an exists.
Then a ∧ a1 = . . . = a ∧ an = 0 and so a ∈ [a1 )0 ∩ . . .∩ [an )0 = ([a1 ) ∩ . . .∩ [an ))0 by 9.
That is a ∈ [a1 ∨. . .∨an )0 . It follows that a∧(a1 ∨. . .∨an ) = 0. Hence a1 ∨. . .∨an ∈
(a)∗ . Thus (a)∗ is an ideal and so S is 0-distributive [see Theorem 2.5, 4].
8 ⇒ 7: Suppose 8 holds and let a, a1 , . . . , an ∈ S be such that [a1 ) ∩ . . . ∩ [an ) 6= ∅.
Then ([a) ∨ ([a1 ) ∩ [a2 )))0 = ([a) ∨ ([a1 ))0 ∩ ([a) ∨ [a2 ))0 . Assume ([a) ∨ ([a1 ) ∩
. . . ∩ [ak−1 )))0 = ([a) ∨ [a1 ))0 ∩ . . . ∩ ([a) ∨ [ak−1 ))0 for 2 < k 6 n. Let x ∈
([a) ∨ [a1 ))0 ∩ . . . ∩ ([a) ∨ [ak ))0 . Then x ∈ ([a) ∨ [a1 ))0 ∩ . . . ∩ ([a) ∨ [ak−1 ))0 =
([a) ∨ ([a1 ) ∩ . . . ∩ [ak−1 )))0 by our induction hypothesis and x ∈ ([a) ∨ [ak ))0 .
Hence x ∧ y = a for some y ∈ [a) ∨ ([a1 ) ∩ . . . ∩ [ak−1 )) and x ∧ z = 0 for some
z ∈ [a) ∨ [ak ). Thus x ∧ a ∧ t = 0 for some t ∈ [a1 ) ∩ . . . ∩ [ak−1 ) and x ∧ a ∧ ak = 0
so that x ∈ [a ∧ t)0 ∩ [a ∧ ak )0 = ([a) ∨ [t))0 ∩ ([a) ∨ [ak ))0 = ([a) ∨ ([t) ∩ [ak )))0
by 8. Consequently x ∧ a ∧ u = 0 for some u ∈ [t) ∩ [ak ) ⊆ [a1 ) ∩ . . . ∩ [ak )
and so x ∈ ([a) ∨ ([a1 ) ∩ . . . ∩ [ak )))0 . Thus ([a) ∨ [a1 ))0 ∩ . . . ∩ ([a) ∨ [ak ))0 ⊆
([a) ∨ ([a1 ) ∩ . . . ∩ [ak )))0 . The reverse inclusion is obvious. By induction it follows
that ([a) ∨ [a1 ))0 ∩ . . . ∩ ([a) ∨ [an ))0 = ([a) ∨ ([a1 ) ∩ . . . ∩ [an )))0 .
Suppose 7 holds. Let a ∈ S and let a1 , . . . , an ∈ (a)∗ be such that a1 ∨ . . . ∨ an
exsits. Then a ∧ a1 = . . . = a ∧ an = 0 and so a ∈ [a1 )0 ∩ . . . ∩ [an )0 . Replacing
a by a1 ∨ . . . ∨ an in 7, we have ([a1 ) ∩ . . . ∩ [an ))0 = [a1 )0 ∩ . . . ∩ [an )0 . Thus
a ∈ ([a1 ) ∩ . . . ∩ [an ))0 = [a1 ∨ . . . ∨ an )0 . Hence a ∧ (a1 ∨ . . . ∨ an ) = 0 and
consequently a1 ∨ . . . ∨ an ∈ (a)∗ . Thus (a)∗ is an ideal. It follows that S is 0distributive [see Theorem 2.5, 4].
!"
2.11. The conditions 3 to 12 of Theorem 2.7 are not necessary for an
infinite semilattice to be 0-distributive. These conditions are both necessary and
sufficient in the case of a lattice.
Clearly each of the conditions 3 to 12 implies the condition 4. Hence it is enough
to prove that 4 is not necessary.
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Let C be an infinite chain without the least element and S = C ∪ {0, a, b, d}.
Define an ordering on S as follows: 0 < a, b, d; akb; akd; bkd and a, b, d < c for all
c ∈ C. Clearly S is a 0-distributive semilattice with respect to this ordering. But no
prime filter of S contains the nonzero element a. Thus 4 is not necessary.
!"
2.12. The conditions 2 to 10 of Theorem 2.8 are not necessary for an
infinite semilattice to be 0-distributive. These conditions are both necessary and
sufficient in the case of a lattice.
Clearly 2 ⇒ 3 ⇒ 4 ⇒ 10, 5 ⇒ 6 ⇒ 8, 7 ⇒ 8, and 9 ⇒ 10. Hence it is enough to
prove that 8 and 10 are not necessary.
Let C be an infinite chain without the least element and S = C ∪ {0, a, b, d, e}.
Define an ordering on S as follows: 0 < a, b, d, e; a < e; akb; akd; bkd; bke; dke;
a, b, d, < c for all c ∈ C; ekc for all c ∈ C. It is easily seen that S is a 0-distributive
semilattice with respect to this ordering. Now [e) ∨ [b) = S = [e) ∨ [d), so that
([e) ∨ [b))0 ∩ ([e) ∨ [d))0 = S. Also [e) ∨ ([b) ∩ [d)) = [a) and hence ([e) ∨ ([b) ∩ [d)))0 =
{0, b, d}. Thus ([e) ∨ ([b) ∩ [d)))0 6= ([e) ∨ [b))0 ∩ ([e) ∨ [d))0 , proving 8 is not necessary.
Consider the 0-distributive semilattice S from Remark 2.11. Now ([a)∩[b))0 = {0}
and [a)0 ∩ [b)0 = {0, d}. Thus ([a) ∩ [b))0 6= [a)0 ∩ [b)0 , proving 10 is not necessary.
!"
2.13. The condition 2 of Theorem 2.7 and the conditions 11, 12,
13 of Theorem 2.8 are necessary for a semilattice (not necessarily finite) to be 0distributive.
.
The necessity of the condition 2 of Theorem 2.7 is obvious. The
necessity of the conditions 11, 12, 13 of Theorem 2.8 follows by Theorem 2.5 [see
Theorem 2.5, 10, 8, 11].
!"
2.14. The condition 2 of Theorem 2.7 and the conditions 11, 12, 13 of
Theorem 2.8 are not sufficient for an infinite semilattice with 0 to be 0-distributive.
Clearly the condition 12 of Theorem 2.8 implies the condition 13 of Theorem 2.8
and the condition 13 of Theorem 2.8 implies the condition 2 of Theorem 2.7. Hence
it is enough to show that the conditions 11 and 12 of Theorem 2.8 are not sufficient.
Let C1 , C2 , C3 be infinite chains without greatest and least elements and let
S = C1 ∪ C2 ∪ C3 ∪ {0, a, b, c, d, e, f, g, 1}. Define an ordering on S as follows. 0 < a,
b, c, d; a < e; b < f ; c < g; d < e; d < f ; d < g; e < c1 < 1 for all c1 ∈ C1 ;
e < c2 < 1 for all c2 ∈ C2 ; f < c1 for all c1 ∈ C1 ; f < c3 < 1 for all c3 ∈ C3 ;
g < c2 for all c2 ∈ C2 ; g < c3 for all c3 ∈ C3 ; akb; akc; akd; akf ; akg; akc3 for all
c3 ∈ C3 ; bkc; bkd; bke; bkg; bkc2 for all c2 ∈ C2 ; ckd; cke; ckf ; ckc1 for all c1 ∈ C1 ;
c1 kc2 for all c1 ∈ C1 and c2 ∈ C2 ; c1 kc3 for all c1 ∈ C1 and c3 ∈ C3 ; c2 kc3 for
all c2 ∈ C2 and c3 ∈ C3 . Clearly S is a semilattice with respect to this ordering.
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Also for all x, y, z ∈ S, we have ((x] ∩ ((y] ∨ (z]))∗ = ((x] ∩ (y])∗ ∩ ((x] ∩ (z])∗ and
((x] ∨ (y]) ∩ ((x] ∨ (z])∗ = (x]∗ ∩ ((y] ∩ (z])∗ . Now (d] ∩ (a] = (0] = (d] ∩ B where
B = (b] ∨ (c]. But (d] ∩ ((a] ∨ B) 6= (0]. Thus S is not 0-distributive.
I would like to thank Prof. P. V. Venkatanarasimhan for his valuable suggestions
in the preparation of this paper. I also thank the referee whose valuable comments
helped in shaping the paper into its present form.
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Author’s address: P. Balasubramani, Department of Mathematics, Kongu Engineering
College, Perundurai, Erode-638 052, India, e-mail: pbalu− 20032001@yahoo.co.in.
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